Optical fiber interconnect having offset ends with reduced tensile stress and fabrication method

ABSTRACT

A method of fabricating a structure comprising an optical fiber interconnect rigidly clamped within an enclosure wherein ends of the interconnect are offset. The method comprises selecting an enclosure material such that its coefficient of thermal expansion leads, during cooling, to a compressive force that minimizes tensile stress in the interconnect. An optical fiber interconnect structure and a semiconductor device fabricated according to the method are also disclosed.

FIELD OF THE INVENTION

The invention relates to optical fiber interconnects having offset ends.

BACKGROUND OF THE INVENTION

Optical fiber interconnects are often subjected to the ends beingoffset. Such an offset can be due to the inability of an employedtechnology to ensure a low enough misalignment of the interconnect ends,or can be an essential feature of a particular photonic package design.If the offset is substantial, it leads to a reactive tensile force inthe fiber. This force occurs because the ends of the interconnect cannotmove closer to compensate for the greater length of the deformed fiber.The reactive tensile force not only results in a tensile stress in thefiber, but also, more importantly, can lead to a significant increase inthe bending stress. The adverse consequences of the tensile stress canbe even greater, if the interconnect is subjected to the additional,“active”, tension. Such a tension can be due, for instance, to thethermal expansion mismatch between the low expansion glass and therelatively high expansion material of the enclosure, when the structureis heated up for reflow soldering, during a laser welding operation, orduring temperature cycling. Elevated tensile stresses and curvatures mayhave an adverse effect on interconnect mechanical strength andtransmission losses. The situation is different, however, in the case ofcompression. Interconnect compression can be caused, for example, by thethermal contraction mismatch of the glass with the material of theenclosure, when the structure is cooled down from a manufacturingtemperature to a lower room or testing temperature.

It is, therefore, desirable to minimize optical fiber interconnectbending to achieve lower tensile stresses and curvatures which mayimprove mechanical strength and minimize transmission losses.

SUMMARY OF THE INVENTION

A method is disclosed of fabricating a structure comprising an opticalfiber interconnect adhesively soldered or epoxy bonded into a ferruleinside an enclosure. An illustrative embodiment of the method comprisesselecting an enclosure material such that a compressive force caused bythe thermal contraction mismatch of the glass fiber with the material ofthe enclosure during the cooling of the structure minimizes tensilestress in the interconnect. Cooling may be for example, from amanufacturing temperature to room temperature.

Further disclosed are an optical fiber interconnect structure and asemiconductor device fabricated according to the method.

DESCRIPTION OF THE DRAWINGS

The invention is best understood from the following detailed descriptionwhen read with the accompanying drawing figures.

FIG. 1. depicts an optical fiber interconnect subjected to an offset ofthe lateral ends and to axial loading.

FIG. 2. depicts auxiliary functions to calculate deflections, rotationalangles and curvatures.

FIG. 3. depicts dimensionless deflection curves for an interconnectsubjected to compression.

FIG. 4. depicts dimensionless curvatures for an interconnect subjectedto compression.

DETAILED DESCRIPTION OF THE INVENTION

A fiber interconnect is typically used at a temperature lower than thatat which it is manufactured. For example, bonding or soldering of theinterconnect to an enclosure may take place at temperatures greater than120° C., whereas the interconnect may be used at room temperature,causing the structure to be subjected to a temperature change of about100° C. Use of interconnects in colder environments causes exposure toeven greater temperature changes. The invention, as it relates to anoptical fiber interconnect with its ends offset, takes advantage ofinduced axial compression arising during structure cooling, to minimizemaximum tensile stress which is due to bending of the interconnect. Thecompression is due to the thermal contraction mismatch of the glass andthe enclosure materials. As used herein, minimizing a stress includesreducing it and eliminating it.

The invention includes a structure comprising a fiber interconnectrigidly clamped within an enclosure, the interconnect having endsoffset, wherein the enclosure material is selected such that itscoefficient of thermal expansion leads, during cooling, to a compressiveforce that minimizes the tensile stress in the interconnect. Thisfavorable compression effect may be achieved through a proper selectionof the enclosure material, i.e. through a selection of an adequatemismatch between the materials of the fiber and the enclosure. When aparticular manufacturing technology is employed, the interconnect couldbe subjected to temperatures exceeding the manufacturing temperatures.As a consequence, the interconnect can experience elevated tensilestresses for a relatively short period of time, before the structure iscooled down to the “steady-state” service temperature. Accordingly,formulas for the prediction of the tensile stresses in a situation, whenthe structure is heated up to a high temperature are also providedherein.

The following analysis examines an optical fiber interconnect subjectedto the ends offset and subsequent axial loading in tension orcompression. Simple analytical stress models are developed for theevaluation of the bending stresses in the interconnect as a result of anot-very-small-ends offset. The term “not-very-small” is used herein toindicate that the fiber deflections, caused by such an off-set, arelarge enough, so that the reactive tension cannot be neglected, but aresmall enough, so that the linear theory of bending of beams can be usedfor the prediction of stresses and strains. An evaluation is presentedof the maximum tensile stress in the fiber due to the reactive tensionand the subsequent thermally induced external (“active”) tension, aswell as to thermal external (“active”) compression. The analyticalmodels are used to design an optical fiber interconnect, characterizedby the minimum possible tensile stress.

1. Optical Fiber Interconnect Subjected to the Ends Offset and ReactiveTension

1.1 Deflection Function

Let an optical fiber interconnect be subjected to a not very small endsoffset, Δ, that results in reactive tensile forces, T₀, as depicted inFIG. 1. The deflection function (elastic curve), w(x), of the fiber canbe determined from the equation of bending

E ₀ I ₀ w ^(IV)(x)−T ₀ w″(x)=0  (1)

and the boundary conditions

w(0)=0, w′(0)=0, w(l)=Δ, w′(l)=0,  (2)

where E₀I₀ is the flexural rigidity of the fiber, E₀ is Young's modulusof the material, I₀=π/4r₀ ⁴ is the moment of inertia of the fibercross-section, r₀ is the radius of the fiber, and l is the interconnectspan. The origin, 0, of the coordinate x is at the left support of theinterconnect. The equation (1), considering the boundary conditions (2),has the following solution: $\begin{matrix}{{{w(x)} = {{\frac{\Delta}{2}1} - {\cosh \quad {kx}} - \frac{\coth \quad {u\left( {{kx} - {\sinh \quad {kx}}} \right)}}{1 - {u\quad \coth \quad u}}}},} & (3)\end{matrix}$

where $\begin{matrix}{k = {\sqrt{\frac{T_{0}}{E_{0}I_{0}}} = {\frac{2}{r_{0}^{2}}\sqrt{\frac{T_{0}}{\pi \quad E_{0}}}}}} & (4)\end{matrix}$

is the parameter of the tensile force, and $\begin{matrix}{u = {\frac{kl}{2} = {\frac{l}{r_{0}^{2}}{\sqrt{\frac{T_{0}}{\pi \quad E_{0}}}.}}}} & (5)\end{matrix}$

is the dimensionless parameter of the compressive force.

If the offset, Δ, is small, the force T₀ is also small, and so are theparameters k and u. Then the following approximate formulas can be used:$\begin{matrix}{{{\coth \quad u} \cong {\frac{1}{u} - \frac{u}{3}}},{{\sinh \quad {kx}} \cong {{kx} + \frac{({kx})^{3}}{6}}},{{\cosh \quad {kx}} \cong {1 + {\frac{({kx})^{2}}{2}.}}}} & (6)\end{matrix}$

Then the solution (3) yields: $\begin{matrix}{{w(x)} = {{w_{0}(x)} = {{\Delta \left( {{3\frac{x^{2}}{l^{2}}} - {2\frac{x^{3}}{l^{3}}}} \right)}.}}} & (7)\end{matrix}$

This formula describes the deflection curve of an interconnect, when theoffset is small, or when the reactive force does not occur.

1.2 Rotating Angles

From the (3) $\begin{matrix}\begin{matrix}{{w^{\prime}(x)} = {{{- \frac{\Delta}{l}}u\sinh \quad {kx}} + \frac{\coth \quad {u\left( {1 - {\cosh \quad {kx}}} \right)}}{1 - {u\quad \coth \quad u}}}} \\{= {{\frac{\Delta}{l}u\cosh \quad u} - \frac{\cosh \left( {{kx} - u} \right)}{{u\quad \cosh \quad u} - {\sinh \quad u}}}} \\{{= {{{w^{\prime}\left( \frac{l}{2} \right)}\cosh \quad u} - \frac{\cosh \left( {{kx} - u} \right)}{{\cosh \quad u} - 1}}},}\end{matrix} & (8)\end{matrix}$

where $\begin{matrix}{{w^{\prime}\left( \frac{l}{2} \right)} = {{{\frac{\Delta}{l}u\cosh \quad u} - \frac{1}{{u\quad \cosh \quad u} - {\sinh \quad u}}} = {\frac{3\Delta}{2l}{\varphi_{0}(u)}}}} & (9)\end{matrix}$

is the angle of rotation at the mid-cross-section of the fiber, and thefunction $\begin{matrix}{{{\varphi_{0}(u)} = {{\frac{2}{3}u\cosh \quad u} - \frac{1}{{u\quad \cosh \quad u} - {\sinh \quad u}}}},{\frac{2}{3} \leq \varphi_{0} \leq 1},} & (10)\end{matrix}$

considers the effect of the tensile force on the angle of rotation atthe mid-cross-section. This function changes from φ₀=1 to φ₀=2/3=0.667,when the tensile force, T₀, changes from zero to infinity. Thecalculated values of the function φ(u) are shown in Table 1 and plottedin FIG. 2.

Table 1. Auxiliary Functions to Calculate Deflections, Rotation Anglesand Curvatures

u 0 π/6 π/3 π/2 2π/3 5π/6 π Φ₀(u) 1.000 0.995 0.983 0.964 0.941 0.9160.892 χ₀(u) 1.000 0.997 0.990 0.980 0.968 0.955 0.944 ψ₀(u) 1.000 1.0181.071 1.154 1.262 1.388 1.528 Φ₀*(u) 1.000 1.005 1.019 1.047 1.095 1.1771.333 χ₀*(u) 1.000 1.003 1.011 1.028 1.059 1.117 1.250 ψ₀*(u) 1.0000.982 0.924 0.822 0.662 0.413 0

1.3 Tensile Strain and Stress

The length of the deflected and stretched fiber can be found as$\begin{matrix}{s = {{\int_{0}^{l}{\sqrt{1 + \left\lbrack {w^{\prime}(x)} \right\rbrack^{2}}\quad {x}}} \cong {l + {\frac{1}{2}{\int_{0}^{l}{\left\lbrack {w^{\prime}(x)} \right\rbrack^{2}\quad {x}}}}}}} & (11)\end{matrix}$

The corresponding tensile strain, ε₀, can be determined, using (8), asfollows: $\begin{matrix}{{ɛ_{0} = {\frac{s - l}{l} = {{\frac{1}{2l}{\int_{0}^{l}{\left\lbrack {w^{\prime}(x)} \right\rbrack^{2}\quad {x}}}} = {\frac{3}{5}\frac{\Delta^{2}}{l^{2}}{\chi_{0}(u)}}}}},} & (12)\end{matrix}$

where the function $\begin{matrix}{{{\chi_{0}(u)} = {\frac{5}{6}u^{2}\frac{1 + {\frac{1}{2}\cosh \quad 2u} - {\frac{3}{2}\frac{\sinh \quad 2\quad u}{2u}}}{\left( {{\sinh \quad u} - {u\quad \cosh \quad u}} \right)^{2}}}},{\chi_{0} \geq 1},} & (13)\end{matrix}$

changes from “one” to infinity, when the tensile force, T₀, changes fromzero to infinity. This function is tabulated in Table 1 and plotted inFIG. 2.

The tensile stress is $\begin{matrix}{\sigma_{t} = {\frac{T_{0}}{\pi \quad r_{0}^{2}} = {E_{0}\frac{r_{0}^{2}}{l^{2}}{u^{2}.}}}} & (14)\end{matrix}$

This stress can be brought down to zero, if the interconnect supportsare moved closer by the distance $\begin{matrix}{\delta_{0} = {{ɛ_{0}l} = {\frac{3}{5}\frac{\Delta^{2}}{l}{\chi_{0}(u)}}}} & (15)\end{matrix}$

to compensate for the reactive tensile strain. If this is done, thedeflection curve will be expressed by the formula (7).

1.4 Bending and Total Stress

The bending stress can be evaluated from the formula (8) as follows:$\begin{matrix}{{{\sigma_{b}(x)} = {{E_{0}r_{0}{w^{''}(x)}} = {{\sigma_{b}(0)}\frac{\sinh \left( {u - {kx}} \right)}{\sinh \quad u}}}},} & (16)\end{matrix}$

where $\begin{matrix}{{\sigma_{b}(0)} = {{- {\sigma_{b}(l)}} = {6E_{0}\frac{\Delta \quad r_{0}}{l^{2}}{\psi_{0}(u)}}}} & (17)\end{matrix}$

is the maximum bending stress that occurs at the end cross-sections, andthe function $\begin{matrix}{{{\psi_{0}(u)} = {\frac{u^{2}}{3}\frac{1}{{u\quad \coth \quad u} - 1}}},{\psi_{0} \geq 1}} & (18)\end{matrix}$

considers the effect of the tensile force on the maximum bending stress.This function changes from “one” to infinity, when the tensile forcechanges from zero to infinity. The function ψ₀(u) is tabulated in Table1 and plotted in FIG. 2.

The total tensile stress in the fiber is due to both tension and bendingand is as follows: $\begin{matrix}{\sigma_{{tot}.} = {{\frac{T_{0}}{\pi \quad r_{0}^{2}} + {\sigma_{b}(0)}} = {{E_{0}\frac{r_{0}^{2}}{l^{2}}u^{2}} + {6E_{0}\frac{\Delta \quad r_{0}}{l^{2}}{{\psi_{0}(u)}.}}}}} & (19)\end{matrix}$

1.5 Effect of the “Active” Tension

The effect of the additional “active” tension, if any, can be evaluatedif a new tensile force, which is expressed as $\begin{matrix}{{T_{1} = {\pi \quad E_{0}r_{0}^{2}\frac{\delta_{0} + \delta_{1}}{l}}},} & (20)\end{matrix}$

is introduced, so that the new parameters of the tensile force are$\begin{matrix}{{k_{1} = {\sqrt{\frac{T_{1}}{E_{0}I_{0}}} = {\frac{2}{r_{0}^{2}}\sqrt{\frac{T_{1}}{\pi \quad E_{0}}}}}},{u_{1} = {\frac{k_{1}l}{2} = {\frac{l}{r_{0}^{2}}{\sqrt{\frac{T_{1}}{\pi \quad E_{0}}}.}}}}} & (21)\end{matrix}$

In the formula (20), δ₁ is the additional, thermally induced,displacement due to the “active” (external) force.

2. Optical Fiber Interconnect Subjected to the Ends Offset and AxialCompression

2.1 Deflection Function

In the case of axial compression, the “minus” sign in front of thesecond term in the equation (1) should be replaced with the “plus” sign.In this case the equation of bending has the following solution:$\begin{matrix}{{{w(x)} = {\frac{\Delta}{2}\frac{1 - {\cos \quad {kx}} - {\cot \quad {u\left( {{kx} - {\sin \quad {kx}}} \right)}}}{1 - {u\quad \cot \quad u}}}},} & (22)\end{matrix}$

The deflection function w(x) is plotted in FIG. 3 for different uvalues. u=π is the condition preceding buckling

From (22) we find: $\begin{matrix}{\left. \begin{matrix}\begin{matrix}{{{w(x)} = {\frac{\Delta}{2}\left( {1 - {\cos \frac{\pi \quad x}{l}}} \right)}},{{{for}\quad u} = \frac{\pi}{2}},} \\{{{w(x)} = {\frac{\Delta}{2}\frac{1 - {\cos \frac{4\pi \quad x}{3l}} - {\frac{1}{\sqrt{3}}\left( {\frac{4\pi \quad x}{3l} - {\sin \frac{4\pi \quad x}{3l}}} \right)}}{1 - \frac{2\pi}{3\sqrt{3}}}}},{{{for}\quad u} = \frac{2\pi}{3}}}\end{matrix} \\{{{w(x)} = {\frac{\Delta}{2\pi}\left( {\frac{2\pi \quad x}{l} - {\sin \frac{2\pi \quad x}{l}}} \right)}},{{{for}\quad u} = \pi}}\end{matrix} \right\}.} & (23)\end{matrix}$

When the force, T₀, is small, the following approximate formulas can beused: $\begin{matrix}{{{\cot \quad u} = {\frac{1}{u} - \frac{u}{3}}},{{\sin \quad {kx}} \cong {{kx} - \frac{({kx})^{3}}{6}}},{{\cos \quad {kx}} \cong {1 - \frac{({kx})^{3}}{2}}},} & (24)\end{matrix}$

and the expression (22) results in the formula (9).

2.2 Rotation Angles

The angles of rotation of the fiber cross-sections can be obtained from(22) by differentiation: $\begin{matrix}{{{w^{\prime}(x)} = {{\frac{\Delta}{l}u\frac{{\cos \left( {{kx} - u} \right)} - {\cos \quad u}}{{\sin \quad u} - {u\quad \cos \quad u}}} = {{w^{\prime}\left( \frac{l}{2} \right)}\frac{{\cos \left( {{kx} - u} \right)} - {\cos \quad u}}{1 - {\cos \quad u}}}}},} & (25)\end{matrix}$

where $\begin{matrix}{{w^{\prime}\left( \frac{l}{2} \right)} = {{\frac{\Delta}{l}u\frac{1 - {\cos \quad u}}{{\sin \quad u} - {u\quad \cos \quad u}}} = {\frac{3\Delta}{2l}{\varphi_{0}^{*}(u)}}}} & (26)\end{matrix}$

is the angle of rotation at the mid-cross-section, and the function$\begin{matrix}{{\varphi_{0}^{*}(u)} = {\frac{2}{3}u\frac{1 - {\cos \quad u}}{{\sin \quad u} - {u\quad \cos \quad u}}}} & (27)\end{matrix}$

consider the effect of the compressive force on the angle of rotation atthe mid-cross-section. The function φ₀*(u) is tabulated in Table 1 andplotted in FIG. 2.

The formulas (26) and (27) yield: $\begin{matrix}{\left. \begin{matrix}\begin{matrix}\begin{matrix}{{{w^{\prime}\left( \frac{l}{2} \right)} = {{\frac{\pi}{2}\frac{\Delta}{l}} = {1.5708\frac{\Delta}{l}}}},{{{for}\quad u} = \frac{\pi}{2}}} \\{{{w^{\prime}\left( \frac{l}{2} \right)} = {{\frac{6\pi}{{3\sqrt{3}} + {2\pi}}\frac{\Delta}{l}} = {1.6420\frac{\Delta}{l}}}},{{{for}\quad u} = \frac{2\pi}{3}}}\end{matrix} \\{{{w^{\prime}\left( \frac{l}{2} \right)} = {{\frac{1 + \sqrt{2}}{1 + \frac{4}{3\pi}}\frac{\Delta}{l}} = {1.6948\frac{\Delta}{l}}}},{{{for}\quad u} = \frac{3\pi}{4}}}\end{matrix} \\{{{w^{\prime}\left( \frac{l}{2} \right)} = {2\frac{\Delta}{l}}},{{{for}\quad u} = \pi}}\end{matrix} \right\}.} & (28)\end{matrix}$

2.3 Compressive Strain

The length, s, of the compressed interconnect (assuming zero initialcompression) is $\begin{matrix}{{s = {{{\int_{0}^{l}{\sqrt{1 - \left\lbrack {w^{\prime}(x)} \right\rbrack^{2}}\quad {x}}} \cong {l - {\frac{1}{2}{\int_{0}^{l}{\left\lbrack {w^{\prime}(x)} \right\rbrack^{2}\quad {x}}}}}} = {l - {\frac{3}{5}\frac{\Delta^{2}}{l^{2}}{\chi_{0}^{*}(u)}}}}},} & (29)\end{matrix}$

where the function $\begin{matrix}{{{\chi_{0}^{*}(u)} = {\frac{5}{6}u^{2}\frac{1 + {\frac{1}{2}\cos \quad 2u} - {\frac{3}{2}\frac{\sin \quad 2u}{2u}}}{\left( {{\sin \quad u} - {u\quad \cos \quad u}} \right)^{2}}}},{\chi_{0}^{*} \geq 1},} & (30)\end{matrix}$

considers the effect of the compressive force. This function istabulated in Table 1 and plotted in FIG. 2. The compressive strain is$\begin{matrix}{ɛ_{0} = {\frac{l - s}{l} = {\frac{3}{5}\frac{\Delta^{2}}{l^{2}}{{\chi_{0}^{*}(u)}.}}}} & (31)\end{matrix}$

The formulas (30) and (31) yield: $\begin{matrix}\left. \begin{matrix}\begin{matrix}\begin{matrix}\begin{matrix}{{ɛ_{0} = {{\frac{3}{5}\frac{\Delta^{2}}{l^{2}}} = {0.6\frac{\Delta^{2}}{l^{2}}}}},{{{for}\quad u} = 0}} \\{{ɛ_{0} = {{\left( \frac{\pi}{4} \right)^{2}\frac{\Delta^{2}}{l^{2}}} = {0.6169\frac{\Delta^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{\pi}{2}}}\end{matrix} \\{{ɛ_{0} = {{2\left( \frac{\pi}{3} \right)^{2}\frac{1 + \frac{3\sqrt{3}}{4\pi}}{\left( {1 + \frac{2\pi}{3\sqrt{3}}} \right)^{2}}\frac{\Delta^{2}}{l^{2}}} = {0.6352\frac{\Delta^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{2\pi}{3}}}\end{matrix} \\{{ɛ_{0} = {{\frac{1 + \frac{1}{\pi}}{\left( {1 + \frac{4}{3\pi}} \right)^{2}}\frac{\Delta^{2}}{l^{2}}} = {0.6498\frac{\Delta^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{3\pi}{4}}}\end{matrix} \\{{ɛ_{0} = {{\frac{3}{4}\frac{\Delta^{2}}{l^{2}}} = {0.75\frac{\Delta^{2}}{l^{2}}}}},{{{for}\quad u} = \pi}}\end{matrix} \right\} & (32)\end{matrix}$

2.4 Bending Stress

Using the expression (25) for the rotation angles, we obtain thefollowing formula for the bending stress: $\begin{matrix}{{{\sigma_{b}(x)} = {{E_{0}r_{0}{w^{''}(x)}} = {{\sigma_{b}(0)}\frac{\sin \left( {u - {kx}} \right)}{\sin \quad u}}}},} & (33)\end{matrix}$

where $\begin{matrix}{{\sigma_{b}(0)} = {{- {\sigma_{b}(l)}} = {6E_{0}\frac{\Delta \quad r_{0}}{l^{2}}{\psi_{0}^{*}(u)}}}} & (34)\end{matrix}$

is the bending stress at the end cross-sections, and the function$\begin{matrix}{{{\psi_{0}^{*}(u)} = {\frac{u^{2}}{3}\frac{1}{1 - {u\quad \cot \quad u}}}},{\psi_{0}^{*} \leq 1}} & (35)\end{matrix}$

considers the effect of the compressive force. This function istabulated in Table 1 and plotted in FIG. 2. The curvature w″(x) isplotted, for different u values, in FIG. 4. The formulas (34) and (35)yield: $\begin{matrix}{\left. \begin{matrix}\begin{matrix}\begin{matrix}\begin{matrix}{{{\sigma_{b}(0)} = {6E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}},{{{for}\quad u} = 0}} \\{{{\sigma_{b}(0)} = {{\frac{\pi^{2}}{2}E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} = {4.9348\frac{\Delta \quad r_{0}}{l^{2}}}}},{{{for}\quad u} = \frac{\pi}{2}}}\end{matrix} \\{{{\sigma_{b}(0)} = {{2\frac{\left( \frac{2\pi}{3} \right)^{2}}{1 + \frac{2\pi}{3\sqrt{3}}}E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} = {3.9711E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}}},{{{for}\quad u} = \frac{2\pi}{3}}}\end{matrix} \\{{{\sigma_{b}(0)} = {{2\frac{\left( \frac{3\pi}{4} \right)^{2}}{1 + \frac{2\pi}{4}}E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} = {3.3083E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}}},{{{for}\quad u} = \frac{3\pi}{3}}}\end{matrix} \\{{{\sigma_{b}(0)} = 0},{{{for}\quad u} = \pi}}\end{matrix} \right\},} & (36)\end{matrix}$

and $\begin{matrix}{\left. \begin{matrix}\begin{matrix}\begin{matrix}\begin{matrix}{{{\sigma_{b}\left( \frac{l}{4} \right)} = {{{\sigma_{b}(0)}\frac{\sin \frac{u}{2}}{\sin \quad u}} = {\frac{\sigma_{b}(0)}{2\cos \frac{u}{b}} = {3E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}}}},{{{for}\quad u} = 0}} \\{{{\sigma_{b}\left( \frac{l}{4} \right)} = {{\frac{\pi^{2}}{2\sqrt{2}}E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} = {3.4894E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}}},{{{for}\quad u} = \frac{\pi}{2}}}\end{matrix} \\{{{\sigma_{b}\left( \frac{l}{4} \right)} = {{\sigma_{b}(0)} = {3.9711E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}}},{{{for}\quad u} = \frac{2\pi}{3}}}\end{matrix} \\{{{\sigma_{b}\left( \frac{l}{4} \right)} = {4.3225E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}},{{{for}\quad u} = \frac{3\pi}{4}}}\end{matrix} \\{{{\sigma_{b}\left( \frac{l}{4} \right)} = {{2\pi \quad E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} = {6.2832E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}}},{{{for}\quad u} = \pi}}\end{matrix} \right\}.} & (37)\end{matrix}$

Comparing the formulas (36) and (37), it may be concluded that when thecompressive force is small, the maximum bending stresses occur at theend cross-sections, and the bending stresses at the points$x = \frac{l}{4}$

and $x = \frac{3l}{4}$

are only half the value of the bending stress at the interconnect ends.When the compressive force increases, the bending stresses at the endcross-sections decrease, and the bending stresses at the cross-sections$x = \frac{l}{4}$

and $x = \frac{3l}{4}$

increase. The stresses at the interconnect ends become equal to thestresses at the cross-sections $x = \frac{l}{4}$

and ${x = \frac{3l}{4}},$

when the compressive force is such that the parameter u becomes$u = {\frac{2\pi}{3} = {2.0944.}}$

Accordingly, compressive forces corresponding to u=2π/3 or less arefavorable. When u=π (the condition preceding buckling), the bendingstresses at the interconnect ends, x=0 and x=l, become zero, and thestresses at the cross-sections $x = \frac{l}{4}$

and $x = \frac{3l}{4}$

are about 4.7% higher than the maximum bending stress at theinterconnect ends in the case of zero compression. $\begin{matrix}{{T_{0} = {{\left( \frac{4\pi}{3} \right)^{2}\frac{E_{0}I_{0}}{l^{2}}} = {1.7778\frac{\pi^{2}E_{0}I_{0}}{l^{2}}}}},} & (38)\end{matrix}$

2.5 Total Stress

The total tensile stress in the interconnect subjected to the endsoff-set and axial compression can be found as $\begin{matrix}{{\sigma_{{tot}.}(x)} = {{{\sigma_{b}(x)} - \frac{T_{0}}{\pi \quad r_{0}^{2}}} = {{{\sigma_{b}(0)}\frac{\sin \left( {u - {kx}} \right)}{\sin \quad u}} - {E_{0}\frac{r_{0}^{2}}{l^{2}}{u^{2}.}}}}} & (39)\end{matrix}$

This formula yields: $\begin{matrix}{{{\sigma_{{tot}.}(0)} = {{\sigma_{b}(0)} - {E_{0}\frac{r_{0}^{2}}{l^{2}}u^{2}}}},{{\sigma_{{tot}.}\left( \frac{l}{4} \right)} = {{\sigma_{b}\left( \frac{l}{4} \right)} - {E_{0}\frac{r_{0}^{2}}{l^{2}}u^{2}}}},} & (40)\end{matrix}$

so that $\begin{matrix}{\left. \begin{matrix}\begin{matrix}\begin{matrix}\begin{matrix}{{{\sigma_{{tot}.}(0)} = {6E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}},{{{for}\quad u} = 0}} \\{{{\sigma_{{tot}.}(0)} = {{4.9348E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} - {2.4674E_{0}\frac{r_{0}^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{\pi}{2}}}\end{matrix} \\{{{\sigma_{{tot}.}(0)} = {{3.9711E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} - {4.3865E_{0}\frac{r_{0}^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{2\pi}{3}}}\end{matrix} \\{{{\sigma_{{tot}.}(0)} = {{3.3083E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} - {5.5517E_{0}\frac{r_{0}^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{3\pi}{4}}}\end{matrix} \\{{{\sigma_{{tot}.}(0)} = {{- 9.8696}E_{0}\frac{r_{0}^{2}}{l^{2}}}},{{{for}\quad u} = \pi}}\end{matrix} \right\}.} & (41)\end{matrix}$

The total tensile stress at the cross-sections $x = \frac{l}{4}$

and $x = \frac{3l}{4}$

is $\begin{matrix}{\left. \begin{matrix}\begin{matrix}\begin{matrix}\begin{matrix}{{{\sigma_{{tot}.}\left( \frac{l}{4} \right)} = {3E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}},{{{for}\quad u} = 0}} \\{{{\sigma_{{tot}.}\left( \frac{l}{4} \right)} = {{3.4894E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} - {2.4674E_{0}\frac{\Delta \quad r_{0}}{l^{2}}}}},{{{for}\quad u} = \frac{\pi}{2}}}\end{matrix} \\{{{\sigma_{{tot}.}\left( \frac{l}{4} \right)} = {{3.9711E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} - {4.3865E_{0}\frac{\Delta \quad r_{0}^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{2\pi}{3}}}\end{matrix} \\{{{\sigma_{{tot}.}\left( \frac{l}{4} \right)} = {{4.3225E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} - {5.5517E_{0}\frac{r_{0}^{2}}{l^{2}}}}},{{{for}\quad u} = \frac{3\pi}{4}}}\end{matrix} \\{{{\sigma_{{tot}.}\left( \frac{l}{4} \right)} = {{6.2832E_{0}\frac{\Delta \quad r_{0}}{l^{2}}} - {9.8696E_{0}\frac{\Delta \quad r_{0}^{2}}{l^{2}}}}},{{{for}\quad u} = \pi}}\end{matrix} \right\}.} & (42)\end{matrix}$

As follows from the expressions (41) and (42), the compressive forcethat corresponds to $u = \frac{\pi}{2}$

results in a compressive stress in the interconnect, if the ratio, Δ/r₀,of the ends offset, Δ, to the fiber radius, r₀, is below 0.5. If thisratio is below 1.10, then a compressive force, corresponding to${u = \frac{2\pi}{3}},$

is required to remove the tensile stress due to bending. In the case of${\frac{\Delta}{r_{0}} < 1.28},$

a force, corresponding to ${u = \frac{3\pi}{4}},$

will be needed. Finally, if the offset-to-radius ratio is below 1.57, acompressive force, corresponding to u=π, will be necessary to eliminatethe tensile stress caused by bending.

For larger offset-to-radius ratios, the total stress in the interconnectwill remain tensile, even if the compressive force will be close to thecritical one. In physical design of the interconnects with ends offset,the favorable effect of compression can be achieved by the properselection of the material of the interconnect enclosure.

The analytical models predict the maximum tensile stress in an opticalfiber interconnect subjected to the ends offset and axial loading intension or compression. If the ends offset is significant, the reactivetension and the would be subsequent external (“active”) tension not onlymay lead to elevated tensile stresses in the fiber, but, moreimportantly, can result in a considerable increase in the bendingstress. Accordingly, the ends offset and the active tension should bereduced in order to minimize the tensile stresses in the fiber. Forcompressive loading there is a way to substantially reduce the initialtensile stress, and, if this stress is not very large, even tosubstantially remove it. This may be done by a proper selection of thecoefficient of thermal expansion (contraction) of the material of theenclosure. Based on the analysis, using the developed stress model, thiscoefficient should be chosen in such a way that the induced compressiveforce in the fiber is about 44% of the critical values. In this case a34% relief in the bending stress, compared to the case of zerocompression can be achieved. A larger compressive force is notadvisable, because of the rapid increase in the fiber curvatures(bending stress) and, as the result of that, in the total maximumtensile stress.

Included in the invention is also a semiconductor device and a method offabricating a semiconductor device. The semiconductor device has anoptical fiber interconnect therein fabricated according to methodsdescribed herein.

Embodiments of the invention include cooling and heating over anytemperature range, provided an enclosure material having a desiredcoefficient of expansion is available. An illustrative temperaturechange is in the range of about 50° C. to about 300° C.

Provided below are examples of applications of analytical models derivedherein. All the numerical examples that follow are carried out for aglass fiber interconnect of the radius r_(o)=0.0625 mm, with thecoefficient of thermal expansion (CTE) of the material ofα_(o)=0.5×10⁻⁶° C.⁻¹, and its Young's modulus of E_(o)=10.5×10⁶ psi=7384kgf/mm². The elevated soldering or curing temperature, if important, isalways assumed to be t_(s)=150° C.

EXAMPLE 1

Determine the critical (Euler) force and stress for a straight opticalfiber interconnect whose length is l=5 mm.

Solution. An optical fiber interconnect, adhesively bonded or solderedat the ends into a capillary or a ferrule can be treated, from thestandpoint of structural analysis, as a single-span beam, rigidlyclamped at the ends. The critical (Euler, or buckling) force for a beamcan be calculated by the formula $\begin{matrix}{T_{e} = \frac{4\pi^{2}E_{0}I_{0}}{l^{2}}} & (1.1)\end{matrix}$

where $\begin{matrix}{I_{0} = {\frac{\pi}{4}r_{0}^{4}}} & (1.2)\end{matrix}$

is the moment of inertia of the fiber cross-section. From (1.1) and(1.2) it follows that:$T_{e} = {\frac{\pi^{3}E_{o}r_{o}^{4}}{l^{2}} = {\frac{\pi^{3} \times 7384 \times 0.0625^{4}}{5^{2}} = {0.1397\quad {kgf}}}}$

The corresponding stress and strain are $\begin{matrix}{\sigma_{e} = {\frac{T_{e}}{\pi \quad r_{o}^{2}} = {\frac{\pi^{2}E_{o}r_{o}^{2}}{l^{2}} = {\frac{\pi^{2} \times 7384 \times 0.0625^{2}}{5^{2}} = {11.3871\quad {{kgf}/{mm}^{2}}}}}}} & (1.3) \\{ɛ_{e} = {\frac{\sigma_{e}}{E_{o}} = {\left( \frac{\pi \quad r_{o}}{l} \right) = {\left( \frac{\pi \times 0.0625}{5} \right)^{2}0.0015421}}}} & (1.4)\end{matrix}$

The distance at which the supports should move closer, so theinterconnect buckles, is $\begin{matrix}{{\Delta \quad l} = {{ɛ_{e}l} = {\frac{\pi^{2}r_{o}^{2}}{l} = {{0.0077106\quad {mm}} = {7.7106\quad {\mu m}}}}}} & (1.5)\end{matrix}$

Note that the buckling force, stress, strain and displacement will notchange, if the interconnect experiences ends offsets of a reasonablemagnitude.

EXAMPLE 2

Determine the CTE of the enclosure that will result in buckling of theinterconnect examined in Example 1.

Solution. The thermally induced stress experienced by the interconnectbecause of its thermal contraction mismatch with the material of theenclosure can be evaluated as

ε₀ =ΔαΔt  (2.1)

where Δt is the change in temperature, and

Δα=α₁−α₀  (2.2)

is the difference in the CTE of the enclosure and the silica materials.Buckling will take place if $\begin{matrix}{{ɛ_{o} \geq ɛ_{e}} = {\left( \frac{\pi \quad r_{o}}{l} \right)^{2} = {\left( \frac{\pi \times 0.0625}{5} \right)^{2} = 0.001542}}} & (2.3)\end{matrix}$

so that the critical value of the CTE of the enclosure is$\begin{matrix}{\alpha_{1}^{e} = {\alpha_{o} + {\frac{1}{\Delta \quad t}\left( \frac{\pi \quad r_{o}}{l} \right)^{2}}}} & (2.4)\end{matrix}$

If the operating temperature is the room temperature (˜25° C.), thenΔt=125° C., and the formula (2.4) yields: $\begin{matrix}{\alpha_{1}^{e} = {{{0.5 \times 10^{- 6}} + {\frac{1}{125}\left( \frac{\pi \times 0.0625}{5} \right)^{2}}} = {12.837 \times 10^{- 6}{^\circ}\quad {C.^{- 1}}}}} & (2.5)\end{matrix}$

If the operating temperature is, for example, 5° C., then α₁^(e)=11.1353×10⁻⁶°C.⁻¹. Hence, if one does not want the interconnect tobuckle, the CTE of the enclosure should not exceed α₁ ^(e)=12.837×10⁻⁶°C.⁻¹, if the operating temperature is 25° C., and α₁ ^(e)=11.135×10⁻⁶°C.⁻¹, if the operating temperature is 5° C.

EXAMPLE 3

The condition which results in an elastically stable fiber cannot bemet. How high is the tensile stress in the buckled fiber? Assume thatthe change in temperature is as large as Δt=180° C.

Solution. Assume that the elastic curve of the buckled interconnect canbe described as $\begin{matrix}{{w(x)} = {f\left( {1 - {\cos 2\pi \quad \frac{x}{l}}} \right)}} & (3.1)\end{matrix}$

This curve satisfies the obvious boundary conditions

w(0)=0, w′(0)=0, w(l),=0, w′(l)=0  (3.2)

The length of the deflected interconnect can be computed as$\begin{matrix}\begin{matrix}{{s(l)} = \quad {{\int_{o}^{l}{\sqrt{1 + \left\lbrack {w^{\prime}(x)} \right\rbrack^{2}}\quad {x}}} \cong {l + {\frac{1}{2}{\int_{o}^{l}{\left\lbrack {w^{\prime}(x)} \right\rbrack^{2}\quad {x}}}}}}} \\{= \quad {{l + {\frac{1}{2}\left( \frac{2\pi \quad f}{l} \right)^{2}{\int_{2}^{l}{\sin^{2}2\pi \quad \frac{x}{l}\quad {x}}}}} = {l + {\left( \frac{\pi \quad f}{l} \right)^{2}{\int_{o}^{l}{\left\lbrack {1 - {\cos 4\pi \quad \frac{x}{l}}} \right\rbrack \quad {x}}}}}}} \\{= \quad {l\left\lbrack {l + \left( \frac{\pi \quad f}{l} \right)^{2}} \right\rbrack}}\end{matrix} & (3.3)\end{matrix}$

The length s(l) should be equal to the initial length, l+δ, of theinterconnect (where δ is the distance, at which the ends of theinterconnect get closer) minus the distance ε_(e)l, because of theelastic compression. Thus, the following condition of the compatibilityof displacements should be fulfilled:

s(l)=l+δ−ε _(e) l  (3.4)

From (3.3) and (3.4) we find: $\begin{matrix}{f = {{\frac{1}{\pi}\sqrt{l\left( {\delta - {ɛ_{e}l}} \right)}} = {\frac{1}{\pi}\sqrt{{l\quad \delta} - \left( {\pi \quad r_{o}} \right)^{2}}}}} & (3.5)\end{matrix}$

From (3.1) we find that the maximum deflection of the buckled fibertakes place at its mid-cross-section (x=l/2): $\begin{matrix}{w_{\max} = {{w\left( \frac{l}{2} \right)} = {2f}}} & (3.6)\end{matrix}$

so that the f value is half the maximum deflection of the buckledinterconnect.

The bending stress in the fiber can be computed as follows:$\begin{matrix}{{\sigma_{B}(x)} = \frac{M(x)}{SM}} & (3.7)\end{matrix}$

where $\begin{matrix}{{M(x)} = {{E_{o}I_{o}{w^{''}(x)}} = {\frac{\pi}{4}E_{o}r_{o}{w^{''}(x)}}}} & (3.8)\end{matrix}$

is the bending moment, and $\begin{matrix}{{SM} = {\frac{\pi}{4}r_{o}^{3}}} & (3.9)\end{matrix}$

is the section modulus of the cross-section. Introducing the formulas(3.8) and (3.9) into the formula (3.7) and considering the expression(3.1), provides: $\begin{matrix}{{\sigma_{b}(x)} = {{E_{o}r_{o}{w^{''}(x)}} = {4\pi^{2}\frac{{fr}_{o}}{l^{2}}E_{o}\cos 2\pi \quad \frac{x}{l}}}} & (3.10)\end{matrix}$

The maximum bending stresses occur at the cross-sections x=0 and x=l atthe interconnect ends, and in its mid-cross-section $x = {\frac{l}{2}:}$

$\begin{matrix}{{\sigma_{b}(0)} = {{\sigma_{b}\left( \frac{l}{2} \right)} = {4\pi^{2}ɛ_{o}\frac{{fr}_{o}}{l^{2}}}}} & (3.11)\end{matrix}$

The compressive stress is equal to the buckling (critical) stress,expressed by the formula (1.3): $\begin{matrix}{\sigma_{c} = {\sigma_{e} = {{E_{o}\left( \frac{\pi \quad r_{o}}{l} \right)}^{2} = {{7384\left( \frac{\pi \times 0.0625}{5} \right)^{2}} = {11.3871\quad {{kgf}\text{/}{mm}^{2}}}}}}} & (3.12)\end{matrix}$

Let, for instance, the CTE of the enclosure be α₁=25×10⁻⁶° C.⁻¹, and theoperating temperature be 5° C., so that the external strain is

ΔαΔt=24.5×10⁻⁶×145=0.0035525  (3.13)

The corresponding axial displacement of the interconnect ends is

δ=lΔαΔt=5×0.0035525=0.0177625 mm,

and the formula (3.5) yields:$f = {{\frac{1}{\pi}\sqrt{{5 \times 0.0177625} - \left( {\pi \times 0.0625} \right)^{2}}} = {0.07136\quad {mm}}}$

The formula (3.11) results in the following bending stress:${\sigma_{b}(0)} = {{4 \times \pi^{2} \times 7384 \times \frac{0.07136 \times 0.0625}{5^{2}}} = {52.006\quad {{kgf}\text{/}{mm}^{2}}}}$

The compressive stress, as defined by the formula (3.12), is$\sigma_{e} = {{7384\left( \frac{\pi \times 0.0625}{5} \right)^{2}} = {11.387\quad {{kgf}\text{/}{mm}^{2}}}}$

Hence, the maximum tensile stress in the buckled fiber is

σ_(t)=σ_(b)(0)−σ_(c)=52.006−11.387=40.619 kgf/mm²

EXAMPLE 4

The optical fiber interconnect is adhesively bonded or soldered at roomtemperature and then its clamped ends are subjected to the ends offset.The fiber is l=5 mm long and the offset is Δ=0.128 mm. Determine thebonding, the tensile and the total maximum stress in tension due to thecombined action of bending and reactive tension.

Solution. The bending stress in the absence of the reactive tension canbe evaluated using the formula (17), in which the function Ψ_(o)(u)should be put equal to one: $\begin{matrix}{\sigma_{b} = {{6E_{o}\Delta \quad \frac{r_{o}}{l^{2}}} = {{6 \times 7384 \times \frac{0.128 \times 0.0625}{5^{2}}} = {14.1773\quad {{kgf}/{mm}^{2}}}}}} & (4.1)\end{matrix}$

The tension strain, ε_(o), can be evaluated on the basis of the formula(12) as $\begin{matrix}{ɛ_{o} = {\frac{3}{5}\frac{\Delta^{2}}{l^{2}}{\chi_{0}(u)}}} & (4.2)\end{matrix}$

and/or on the basis of the formula (14) as $\begin{matrix}{ɛ_{o} = {\frac{\sigma_{t}}{E_{o}} = \left( \frac{r_{o}u}{l} \right)^{2}}} & (4.3)\end{matrix}$

Comparing the formulas (4.2) and (4.3), we conclude that the u value(parameter of the tensile force) can be found from the equation$\begin{matrix}{\frac{u^{2}}{\chi_{2}(u)} = {{\frac{3}{5}\frac{\Delta^{2}}{r_{0}^{2}}} = 2.5166}} & (4.4)\end{matrix}$

Using the Table 1 data, we find that if the u value is equal to${u = {\frac{\pi}{2} = 1.5708}},$

then χ_(o)(u)=0.980, and the ratio u²/χ(u) is 2.5178, which is insatisfactory agreement with the result (4.4). From the Table 1 we have:ψ₀(u)=1.154. The bending stress, with consideration of the effect oftension can be evaluated the formula (17): $\begin{matrix}{\sigma_{b} = {{6E_{o}\frac{\Delta \quad r_{o}}{l^{2}}{\psi_{o}(u)}} = {{14.1773 \times 1.154} = {16.3606\quad {{kgf}\text{/}{mm}^{2}}}}}} & (4.5)\end{matrix}$

The tensile stress can be found from the formula (14): $\begin{matrix}{\sigma_{t} = {{E_{o}\left( \frac{r_{o}u}{l} \right)}^{2} = {{7384\left( \frac{0.0625 \times 1.5708}{5} \right)^{2}} = {2.8468\quad {{kgf}\text{/}{mm}^{2}}}}}} & (4.6)\end{matrix}$

Hence, the total tensile stress is

σ_(tot.)=σ_(b)+σ_(t)=19.2074 kgf/mm²  (4.7)

The reactive tension can be relieved if the interconnect supports moveclose at the distance $\begin{matrix}{\delta_{o} = {{ɛ_{o}l} = {{\left( \frac{r_{o}u}{l} \right)^{2}l} = {{\left( \frac{0.0625 \times 1.5708}{5} \right)^{2}5} = {{0.0019277\quad {mm}} = {1.9277\quad {\mu m}}}}}}} & (4.8)\end{matrix}$

The corresponding strain is

ε_(o)=0.00038554  (4.9)

EXAMPLE 5

Al=5 mm long initially straight interconnect is bonded at its ends intoa capillary at the temperature of 150° C., then cooled to the roomtemperature of 25° C., and then subjected to the ends offset of Δ=0.128mm. The CTE of the enclosure in α₁=10×10⁻⁶° C.⁻¹. Does the fiberexperience tensile stress?

Solution. The compressive strain in the fiber at room temperature,before it becomes subjected to the ends offset, is

ε_(o) =ΔαΔt=9.5×10⁻⁶×125=0.0011875  (5.1)

The tensile strain , as predicted by the formula (4.9), isε_(o)=0.00038554, so that the axial compression in the fiber results inthe compressive stress of

σ_(c)=7384(0.0011875−0.00038554)=5.9217 kgf/mm²

The compressive force is

T _(c)=σ_(c) πr ₀ ²=5.9217×π×0.0625²=0.07267 kgf

The parameter of the axial compressive force is$u = {{\frac{l}{r_{o}^{2}}\sqrt{\frac{T_{c}}{\pi \quad E_{o}}}} = {{\frac{5}{0.0625^{2}}\sqrt{\frac{0.07267}{\pi \times 7384}}} = 2.2655}}$

The Table 1 yields: χ_(o)*(u)=1.078. Then the bending stress, inaccordance with the formula (34), is${\sigma_{b}(0)} = {{6E_{o}\frac{\Delta \quad r_{o}}{l^{2}}{\psi_{o}^{*}(u)}} = {{14.1773 \times 1.078} = {15.2825\quad {{kgf}\text{/}{mm}^{2}}}}}$

The tensile stress in the fiber is

σ_(t)=σ_(b)(0)−σ_(c)=15.2825−5.9217=9.3608 kgf/mm²

EXAMPLE 6

The fiber interconnect examined in the previous examples is subjectedfirst to the offset of Δ=0.128 mm, then the structure is heated up forbonding to the temperature of 150° C. and epoxy material in thecapillary is cured at this temperature. Then the fiber is cooled down toroom temperature. Select the CTE of the enclosure in such a way that thetensile stress in the fiber is minimized or, if possible, eliminated.

Solution. When the fiber is subjected to the ends offset, it experiencesbending stresses only, and the maximum bonding stress is

σ_(b)=14.1773 kgf/mm²  (6.1)

Since the fiber is allowed to slip in at least one of the supports, noadditional stresses occur, when the fiber is heated up to 150° C.However, when it is cooled to room temperature, it experiencescompression, and the external compressive strain is

ε_(o)=ΔαΔτ  (6.2)

The corresponding compressive force is

T _(c) =πr ₀ ² E ₀ΔαΔt  (6.3)

and the parameter of the compressive force can be computed as$\begin{matrix}{u = {{\frac{l}{r_{o}^{2}}\sqrt{\frac{T_{e}}{\pi \quad E_{o}}}} = {\frac{l}{r_{o}}\sqrt{\Delta \quad \alpha \quad \Delta \quad t}}}} & (6.4)\end{matrix}$

The study suggests that the u value be chosen, for the lowest totalstress in tension, as $u = {\frac{2\pi}{3}.}$

Then the formula (6.4) yields:${\Delta \quad \alpha} = {{\frac{1}{\Delta \quad t}\left( \frac{2\pi}{3l} \right)^{2}} = {{\frac{1}{125}\left( \frac{2 \times \pi \times 0.0625}{3 \times 5} \right)^{2}} = {5.4831 \times 10^{- 6}{^\circ}\quad {C^{- 1}.}}}}$

Then the CTE of the enclosure should be${\Delta \quad \alpha} = {{\frac{1}{\Delta \quad t}\left( \frac{2\pi \quad r_{0}}{3l} \right)^{2}} = {{\frac{1}{125}\left( \frac{2 \times \pi \times 0.0625}{3 \times 5} \right)^{2}} = {5.4831 \times 10^{- 6}{^\circ}\quad {C^{- 1}.}}}}$

This coefficient of thermal expansion is or near that of Kovar™. Thus, aKovar™ enclosure will minimize tensile stress in the fiber when thestructure is cooled from 150 ° C. to room temperature. The total tensilestress, predicted by the third formula in (42) is $\begin{matrix}{\sigma_{{tot}.} = \quad {{3.9711E_{0}\quad \frac{\Delta \quad r_{0}}{l^{2}}} - {4.3865E_{o}\quad \frac{r_{o}}{l^{2}}}}} \\{= \quad {{3.9711 \times 7384 \times \frac{0.0625 \times 0.128}{5^{2}}} - {4.3865 \times 7384 \times \frac{0.0625^{2}}{5^{2}}}}} \\{= \quad {4.3223\quad {{kgf}/{mm}^{2}}}}\end{matrix}$

This stress is significantly lower than the “initial” bending stress of14.1773 kgf/mm².

While the invention has been described by illustrative embodiments,additional advantages and modifications will occur to those skilled inthe art. Therefore, the invention in its broader aspects is not limitedto specific details shown and described herein. Modifications, forexample, to structure dimensions, materials and uses, and tocalculations of desirable coefficients of thermal expansion, may be madewithout departing from the spirit and scope of the invention.Accordingly, it is intended that the invention not be limited to thespecific illustrative embodiments but be interpreted within the fullspirit and scope of the appended claims.

What is claimed is:
 1. A method of fabricating a structure comprising anoptical fiber interconnect rigidly clamped within an enclosure whereinends of the interconnect are offset, the method comprising: selecting anenclosure material such that its coefficient of thermal expansion leads,during cooling, to a compressive force on the interconnect thatminimizes tensile stress in the interconnect.
 2. The method of claim 1wherein the structure is cooled from a manufacturing temperature to roomtemperature.
 3. The method of claim 1 wherein the structure is cooled byan amount in the range of about 50° C. to about 300° C.
 4. The method ofclaim 1 wherein the compressive force corresponds to a compressive forceparameter, u, less than or equal to about 2π/3.
 5. The method of claim 1wherein the compressive force is such that stresses at the ends of theinterconnect are substantially equal to stresses at fiber cross-sectionsat distances, x=l/4 and x=3l/4 from the interconnect ends wherein l isthe interconnect span.
 6. The method of claim 1 wherein the compressiveforce is less than or equal to about 44% of the critical force.
 7. Themethod of claim 6 wherein the compressive force is in the range of about35% to about 44% of the critical force.
 8. A structure comprising anoptical fiber interconnect rigidly clamped within an enclosure, theoptical fiber interconnect having ends offset, wherein the enclosurematerial has a coefficient of thermal expansion such that a compressiveforce on the interconnect caused by cooling the structure minimizes themaximum tensile stress in the interconnect.
 9. The structure of claim 8wherein the tensile stress is reduced as the structure is cooled from amanufacturing temperature to room temperature.
 10. The structure ofclaim 8 wherein the compressive force corresponds to a compressive forceparameter, u, less than or equal to about 2π/3.
 11. The structure ofclaim 8 wherein the compressive force is such that stresses at the endsof the interconnect are substantially equal to stresses at fibercross-sections at distances, x=l/4 and x=3l/4 from an interconnect endwherein l is the interconnect span.
 12. The structure of claim 8 whereinthe compressive force is less than or equal to about 44% of the criticalforce.
 13. The structure of claim 12, wherein the compressive force isin the range of about 35% to about 44% of the critical force.
 14. Asemiconductor device having an optical fiber interconnect therein,wherein the optical fiber interconnect is fabricated according to themethod in claim
 1. 15. A method of fabricating a semiconductor devicehaving an optical fiber interconnect therein, wherein the optical fiberinterconnect is fabricated according to claim
 1. 16. A structurecomprising an optical fiber interconnect having ends rigidly clampedwithin an enclosure, the ends of the interconnect being offset, whereinthe enclosure material has a coefficient of thermal expansion such thatan axial compression of the interconnect caused by cooling the structureminimizes the maximum tensile stress in the interconnect.
 17. Thestructure of claim 16, wherein the axial compression is such thatstresses at the ends of the interconnect are substantially equal tostresses at fiber cross-sections at distances x=l/4 and x=3l/4 from aninterconnect end wherein l is the interconnect span.